Fourier Series Integration
📂Fourier Analysis Fourier Series Integration Theorem Let’s say a periodic function f f f 2 L 2L 2 L piecewise continuous in the interval [ − L , L ) [-L,\ L) [ − L , L ) f f f
∫ t 1 t 2 f ( t ) d t = c 0 ( t 2 − t 1 ) + ∑ n ≠ 0 L i n π c n ( e i n π t 2 L − e i n π t 1 L )
\int_{t_{1}}^{t_{2}} f(t) dt= c_{0}(t_{2}-t_{1}) +\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n}\left( e^{i\frac{n\pi t_{2}}{L}}-e^{i\frac{n\pi t_{1}}{L}} \right)
∫ t 1 t 2 f ( t ) d t = c 0 ( t 2 − t 1 ) + n = 0 ∑ inπ L c n ( e i L nπ t 2 − e i L nπ t 1 )
Here, c 0 , c n c_{0},\ c_{n} c 0 , c n complex Fourier coefficient .
Thus, the definite integral of f ( t ) f(t) f ( t ) f ( t ) f(t) f ( t )
Proof Let’s define a function F F F
F ( t ) : = ∫ 0 t f ( s ) d s − c 0 t
F(t):= \int_{0}^t f(s)ds-c_{0}t
F ( t ) := ∫ 0 t f ( s ) d s − c 0 t
Then, it can be shown that F F F 2 L 2L 2 L
F ( t + 2 L ) = ∫ 0 t + 2 L f ( s ) d s − c 0 ( t + 2 L ) = ∫ 0 t f ( s ) d s + ∫ t 2 L f ( s ) d s − c 0 t − 2 L c 0 = ∫ 0 t f ( s ) d s − c 0 t = F ( t )
\begin{align*}
F(t+2L) &= \int_{0}^{t+2L} f(s)ds - c_{0}(t+2L)
\\ &= \int_{0}^t f(s)ds+\int_{t}^{2L}f(s)ds - c_{0}t -2Lc_{0}
\\ &= \int_{0} ^t f(s)ds-c_{0}t=F(t)
\end{align*}
F ( t + 2 L ) = ∫ 0 t + 2 L f ( s ) d s − c 0 ( t + 2 L ) = ∫ 0 t f ( s ) d s + ∫ t 2 L f ( s ) d s − c 0 t − 2 L c 0 = ∫ 0 t f ( s ) d s − c 0 t = F ( t )
Since f f f F F F F F F Dirichlet conditions and can be expanded into a Fourier series.
F ( t ) = C 0 + ∑ n ≠ 0 C n e i n π L t
F(t)=C_{0}+\sum \limits_{n \ne 0} C_{n}e^{i\frac{n\pi}{L}t}
F ( t ) = C 0 + n = 0 ∑ C n e i L nπ t
The reason for separating the constant term will be known later. When calculating the coefficients,
C n = 1 2 L ∫ − L L F ( t ) e − i n π L t d t = − L i n π 1 2 L [ F ( t ) e i n π L t ] − L L + L i n π 1 2 L ∫ − L L F ′ ( t ) e − i n π L t d t = L i n π 1 2 L ∫ − L L ( f ( t ) − c 0 ) e − i n π L t d t = L i n π 1 2 L ∫ − L L f ( t ) e − i n π L t d t − L i n π c 0 2 L ∫ − L L e − i n π L t d t = L i n π 1 2 L ∫ − L L f ( t ) e − i n π L t d t = L i n π c n
\begin{align*}
C_{n}&=\dfrac{1}{2L}\int_{-L}^{L} F(t)e^{-i\frac{n\pi}{L}t} dt
\\ &= \dfrac{-L}{in\pi} \dfrac{1}{2L} \left[ F(t)e^{i\frac{n\pi}{L}t} \right]_{-L}^{L} +\dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} F^{\prime}(t)e^{-i\frac{n\pi }{L}t}dt
\\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} \left( f(t)-c_{0} \right) e^{-i\frac{n\pi}{L}t}dt
\\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} f(t)e^{-i\frac{n\pi}{L}t}dt -\dfrac{L}{in\pi}\dfrac{c_{0}}{2L} \int_{-L}^{L} e^{-i\frac{n\pi}{L}t} dt
\\ &= \dfrac{L}{in\pi}\dfrac{1}{2L} \int_{-L}^{L} f(t)e^{-i\frac{n\pi}{L}t}dt
\\ &= \dfrac{L}{in\pi}c_{n}
\end{align*}
C n = 2 L 1 ∫ − L L F ( t ) e − i L nπ t d t = inπ − L 2 L 1 [ F ( t ) e i L nπ t ] − L L + inπ L 2 L 1 ∫ − L L F ′ ( t ) e − i L nπ t d t = inπ L 2 L 1 ∫ − L L ( f ( t ) − c 0 ) e − i L nπ t d t = inπ L 2 L 1 ∫ − L L f ( t ) e − i L nπ t d t − inπ L 2 L c 0 ∫ − L L e − i L nπ t d t = inπ L 2 L 1 ∫ − L L f ( t ) e − i L nπ t d t = inπ L c n
Therefore,
F ( t ) = C 0 + ∑ n ≠ 0 L i n π c n e i n π L t
F(t)=C_{0} +\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t}
F ( t ) = C 0 + n = 0 ∑ inπ L c n e i L nπ t
Using the definition of F ( t ) F(t) F ( t ) f ( t ) f(t) f ( t )
∫ t 1 t 2 f ( t ) d t = ∫ 0 t 2 f ( t ) d t − ∫ 0 t 1 f ( t ) d t = F ( t 2 ) − F ( t 1 ) + c 0 ( t 2 − t 1 ) = ( C 0 + ∑ n ≠ 0 L i n π c n e i n π L t 2 ) − ( C 0 + ∑ n ≠ 0 L i n π c n e i n π L t 1 ) + c 0 ( t 2 − t 1 ) = c 0 ( t 2 − t 1 ) + ∑ n ≠ 0 L i n π c n ( e i n π L t 2 − e i n π L t 1 )
\begin{align*}
\int_{t_{1}}^{t_{2}} f(t) dt &= \int_{0}^{t_{2}} f(t) dt -\int_{0}^{t_{1}} f(t)dt
\\ &= F(t_{2})- F(t_{1}) + c_{0}(t_{2}-t_{1})
\\ &= \left( C_{0}+\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t_{2}} \right) -\left( C_{0}+\sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} e^{i\frac{n\pi}{L}t_{1}} \right) +c_{0}(t_{2}-t_{1})
\\ &= c_{0}(t_{2}-t_{1}) + \sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} \left( e^{i\frac{n\pi}{L}t_{2}}-e^{i\frac{n\pi}{L}t_{1}} \right)
\end{align*}
∫ t 1 t 2 f ( t ) d t = ∫ 0 t 2 f ( t ) d t − ∫ 0 t 1 f ( t ) d t = F ( t 2 ) − F ( t 1 ) + c 0 ( t 2 − t 1 ) = C 0 + n = 0 ∑ inπ L c n e i L nπ t 2 − C 0 + n = 0 ∑ inπ L c n e i L nπ t 1 + c 0 ( t 2 − t 1 ) = c 0 ( t 2 − t 1 ) + n = 0 ∑ inπ L c n ( e i L nπ t 2 − e i L nπ t 1 )
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As mentioned at the beginning of the document, note that the series on the right is not the Fourier series of ∫ t 1 t 2 f ( t ) d t \int_{t_{1}}^{t_{2}}f(t)dt ∫ t 1 t 2 f ( t ) d t ∫ t 1 t 2 f ( t ) d t − c 0 ( t 2 − t 1 ) \displaystyle \int_{t_{1}}^{t_{2}} f(t) dt -c_{0}(t_{2}-t_{1}) ∫ t 1 t 2 f ( t ) d t − c 0 ( t 2 − t 1 ) ∑ n ≠ 0 L i n π c n ( e i n π L t 2 − e i n π L t 1 ) \sum \limits_{n \ne 0} \dfrac{L}{in\pi}c_{n} \left( e^{i\frac{n\pi}{L}t_{2}}-e^{i\frac{n\pi}{L}t_{1}} \right) n = 0 ∑ inπ L c n ( e i L nπ t 2 − e i L nπ t 1 )
